\(^{\dfrac{2008a-2009b}{2009c+2010d}}=\dfrac{2008c-2009d}{2009a+2010b}.\dfrac{b^2}{d^2}\)
Làm hộ mk với !!! HELP ME !!!!
Cho \(\dfrac{a}{b}=\dfrac{b}{c}.CMR:\dfrac{2008a-2009b}{2009c+2010d}=\dfrac{2008c-2009d}{2009a+2010d}\)
Ta có : \(\dfrac{a}{b}=\dfrac{c}{d}\)
➩\(\dfrac{a}{c}=\dfrac{b}{d}\)➩\(\dfrac{2008a}{2009c}=\dfrac{2009b}{2010d}=\dfrac{2008a+2009b}{2009c+2010d}=\dfrac{2008a-2009b}{2009c-2010}\)
➩\(\dfrac{2008a-2009b}{2009c+2009c}=\dfrac{2008c-2009d}{2009a+2010d}\left(đpcm\right)\)
* đpcm : điều phải chứng minh
Chúc bạn học tốt !!!
Nếu thấy đúng thì tick cho mình nhé !!!
\(Cho\frac{a}{b}=\frac{b}{c}.CMR:\frac{2008a-2009b}{2009c+2010d}=\frac{2008c-2009d}{2009a+2010d}\)
Bài 2 cho tỉ lệ a/b=c/d
a, ab/cd=(a-b)mũ 2/ (c-d)mũ 2
b, (a+b/c+d)= a mũ 2+5d/ c mũ 2+d mũ 2
c,a-b/a+b=c-d/c+d
d,2a+5b/3a-4b=2c+5d/3c-4d
e,2008a-2009b/2009c+2010d=2008c-2009d/2009a+2010b
giúp mk nha ai nhanh mk tick cho
bài 1 tìm x
3 nhân (3-x)-4 nhân l1-xl=15
Bài2 Tính số đo 3 góc trongg của một hình tam giác biết 3 góc tỉ lệ với 2,4,3
bài 3 cho tỉ lệ thức a/b=cd
chứng minh ta có các tir lệ thức:
a)( (a+b/c+d)2=a2+b2/c2-d2
b, a-/a+b=c-d/c+d
c, ab/cd=(a-b)2/(c-d)2
d, 2a+5b/3a-4b=2c+5d/3c-4d
e,2008a-2009b/2009c+2010d=2008c-2009d/2009a+2010b
giúp mk nha nhanh mk tick
Bài 2:
Áp dụng tính chất của dãy tỉ số bằng nhau, ta được:
\(\dfrac{a}{2}=\dfrac{b}{4}=\dfrac{c}{3}=\dfrac{a+b+c}{2+4+3}=\dfrac{180}{9}=20\)
=>a=20; b=80; c=60
Bài 3:
a: Đặt a/b=c/d=k
=>a=bk; c=dk
\(\left(\dfrac{a+b}{c+d}\right)^2=\left(\dfrac{bk+b}{dk+d}\right)^2=\dfrac{b^2}{d^2}\)
\(\dfrac{a^2-b^2}{c^2-d^2}=\dfrac{b^2k^2-b^2}{d^2k^2-d^2}=\left(\dfrac{b}{d}\right)^2\)
Do đó: \(\left(\dfrac{a+b}{c+d}\right)^2=\dfrac{a^2-b^2}{c^2-d^2}\)
c: \(\dfrac{ab}{cd}=\dfrac{bk\cdot b}{dk\cdot d}=\dfrac{b^2}{d^2}\)
\(\left(\dfrac{a-b}{c-d}\right)^2=\left(\dfrac{bk-b}{dk-d}\right)^2=\dfrac{b^2}{d^2}\)
Do đó: \(\dfrac{ab}{cd}=\left(\dfrac{a-b}{c-d}\right)^2\)
cho tỉ lệ thức: a/b=c/d. CMR: ac/bd = 2009a^2+2010c^2/ 2009b^2+2010d^2
\(\frac{a}{b}=\frac{c}{d}\)
\(\Rightarrow\frac{a^2}{b^2}=\frac{c^2}{d^2}=\frac{a}{b}.\frac{c}{d}=\frac{ac}{bd}\)
\(\Rightarrow\frac{2009a^2}{2009b^2}=\frac{2010c^2}{2010d^2}=\frac{2009a^2+2010c^2}{2009b^2+2010d^2}=\frac{ac}{bd}\)
Cho \(\dfrac{a}{2b}=\dfrac{b}{2c}=\dfrac{c}{2d}=\dfrac{d}{2a}\left(a,b,c,d>0\right)\)
Tính: \(\dfrac{2011a-2010b}{c+d}+\dfrac{2011b-2010c}{a+d}+\dfrac{2011c-2010d}{a+b}+\dfrac{2011d-2010a}{b+c}\)
Từ \(\dfrac{a}{2b}=\dfrac{b}{2c}=\dfrac{c}{2d}=\dfrac{d}{2a}\Rightarrow\dfrac{1}{2}.\dfrac{a}{b}=\dfrac{1}{2}.\dfrac{b}{c}=\dfrac{1}{2}.\dfrac{c}{d}=\dfrac{1}{2}.\dfrac{d}{a}\)
⇒ \(\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{d}=\dfrac{d}{a}=\dfrac{a+b+c+d}{b+c+d+a}=1\)
⇒ \(a=b=c=d\)
Thay b = a ; c = a ; d = a vào biểu thức A ta có:
\(A=\dfrac{2011a-2010a}{2a}+\dfrac{2011a-2010a}{2a}+\dfrac{2011a-2010a}{2a}+\dfrac{2011a-2010a}{2a}\)
\(A=\dfrac{a}{2a}+\dfrac{a}{2a}+\dfrac{a}{2a}+\dfrac{a}{2a}\)
\(A=\dfrac{1}{2}.4=2\)
Vậy A = 2
\(\dfrac{a}{2b}=\dfrac{b}{2c}=\dfrac{c}{2d}=\dfrac{d}{2a}\)
Áp dụng tính chất của dãy tỉ số bằng nhau, ta có:
\(\dfrac{a}{2b}=\dfrac{b}{2c}=\dfrac{c}{2d}=\dfrac{d}{2a}=\dfrac{a+b+c+d}{2a+2b+2c+2d}=\dfrac{1}{2}\)
=>\(\dfrac{a}{2b}=\dfrac{1}{2}\)=>2a=2b =>a=b
\(\dfrac{b}{2c}=\dfrac{1}{2}\)=>2b=2c =>b=c
\(\dfrac{c}{2d}=\dfrac{1}{2}\)=>2c=2d =>c=d
\(\dfrac{d}{2a}=\dfrac{1}{2}\)=>2d=2a =>d=a
=>a=b=c=d.
*\(\dfrac{2011a-2010b}{c+d}+\dfrac{2011b-2010c}{a+d}+\dfrac{2011c-2010d}{a+b}+\dfrac{2011d-2010a}{b+c}\)
=\(\dfrac{2011a-2010a}{a+a}+\dfrac{2011a-2010a}{a+a}+\dfrac{2011a-2010d}{a+a}+\dfrac{2011a-2010a}{a+a}\)
=\(\dfrac{a}{2a}+\dfrac{a}{2a}+\dfrac{a}{2a}+\dfrac{a}{2a}\)=2
Bài 2 : Cho \(\dfrac{a}{2b}=\dfrac{b}{2c}=\dfrac{c}{2d}=\dfrac{d}{2a}\left(a,b,c,d>0\right)\)
Tính
\(A=\dfrac{2011a-2010b}{c+d}+\dfrac{2011b-2010c}{a+d}+\dfrac{2011c-2010d}{a+b}+\dfrac{2011d-2010a}{b+c}\)
Lời giải:
Áp dụng tính chất dãy tỉ số bằng nhau:
\(\frac{a}{2b}=\frac{b}{2c}=\frac{c}{2d}=\frac{d}{2a}=\frac{a+b+c+d}{2b+2c+2d+2a}=\frac{a+b+c+d}{2(a+b+c+d)}=\frac{1}{2}\)
\(\Rightarrow \frac{a}{b}=\frac{b}{c}=\frac{c}{d}=\frac{d}{a}=1\Leftrightarrow a=b=c=d\)
Do đó:
\(A=\frac{2011a-2010a}{a+a}+\frac{2011a-2010a}{a+a}+\frac{2011a-2010a}{a+a}+\frac{2011a-2010a}{a+a}\)
\(\Leftrightarrow A=\frac{1}{2}+\frac{1}{2}+\frac{1}{2}+\frac{1}{2}=2\)
Vậy \(A=2\)
Ta có: \(\dfrac{a}{2b}=\dfrac{b}{2c}=\dfrac{c}{2d}=\dfrac{d}{2a}=\dfrac{a+b+c+d}{2\left(a+b+c+d\right)}=\dfrac{1}{2}\)
\(\Rightarrow a=b;b=c;c=d;d=a\)
\(A=\dfrac{2011a-2010b}{c+d}+\dfrac{2011b-2010c}{a+d}+\dfrac{2011c-2010d}{a+b}+\dfrac{2011d-2010a}{b+c}\)
\(A=\dfrac{2011c-2010c}{c+c}+\dfrac{2011c-2010c}{c+c}+\dfrac{2011c-2010c}{c+c}+\dfrac{2011c-2010c}{c+c}\)
\(A=\dfrac{c+c+c+c}{c+c}=2\)
Vậy ....................
cho \(\dfrac{a}{2b}=\dfrac{b}{2c}=\dfrac{c}{2d}=\dfrac{d}{2a}\)
(a,b,c,d >0)
tính A\(=\dfrac{2011a-2010b}{c+d}+\dfrac{2011b-2010c}{a+d}+\dfrac{2011c-2010d}{a+b}+\dfrac{2011d-2011a}{b+c}\)
theo đề bài, \(\dfrac{a}{2b}=\dfrac{b}{2c}=\dfrac{c}{2d}=\dfrac{d}{2a}\)=> a=b=c=d và lớn hơn 0
thay b,c,d thành a, có:
A= \(\dfrac{2011a-2010b}{c+d}\)\(+\dfrac{2011b-2010c}{a+d}+\dfrac{2011c-2010d}{a+b}+\dfrac{2011d-2011a}{b+c}\)
=\(\dfrac{a\left(2011-2010\right)}{2a}+\dfrac{a\left(2011-2010\right)}{2a}+\dfrac{a\left(2011-2010\right)}{2a}+\dfrac{a\left(2011-2011\right)}{2a}\)
=\(\dfrac{1}{2}+\dfrac{1}{2}+\dfrac{1}{2}+0=\dfrac{3}{2}=1.5\)
Cho\(\dfrac{a}{2b}=\dfrac{b}{2c}=\dfrac{c}{2d}=\dfrac{d}{2a}\)
tính P biết
\(\dfrac{2011a-2010b}{c+d}+\dfrac{2011b-2010c}{a+d}+\dfrac{2011c-2010d}{a+b}+\dfrac{2011d-2010a}{b+c}\)
\(\dfrac{a}{2b}=\dfrac{b}{2c}=\dfrac{c}{2d}=\dfrac{d}{2a}=\dfrac{a+b+c+d}{2\left(a+b+c+d\right)}=\dfrac{1}{2}\)
( theo tính chất dãy tỉ số bằng nhau )
\(\Rightarrow\left\{{}\begin{matrix}a=\dfrac{1}{2}\cdot2b\\b=\dfrac{1}{2}\cdot2c\\c=\dfrac{1}{2}\cdot2d\\d=\dfrac{1}{2}\cdot2a\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=b\\b=c\\c=d\\d=a\end{matrix}\right.\Rightarrow a=b=c=d\)
\(\Rightarrow P=\dfrac{a}{2a}+\dfrac{a}{2a}+\dfrac{a}{2a}+\dfrac{a}{2a}=2\)